$\int \dfrac{x^2}{x^3+4}\,dx\,= $ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac13\ln|x^3+4|+C$ (Choice B) B $\ln|x|+\dfrac{x^3}{12}+C$ (Choice C) C $\ln|x^3+4|\cdot\dfrac{x^3}{3}+C$ (Choice D) D $\ln\left|\dfrac{x^4}{4}+4x\right|+C$
Explanation: Notice that we can rewrite the integral as $ \int \dfrac{1}{x^3+4}\,\cdot x^2\,dx\,$. If we let $ {u=x^3+4}$, then ${du=3x^2 \, dx}$ and $\dfrac{du}{3}=x^2 \, dx} $ Substituting gives us: $\begin{aligned} \int \dfrac{1}{{x^3+4}}\,\cdot x^2\,dx}\, &= \int \dfrac{1}{ u}\,\cdot \dfrac{du}{3}}\\\\\\ &= \int \dfrac{1}{\ u}\,\cdot \dfrac13\,du\\\\\\ &=\dfrac13 \int \dfrac{1}{\ u}\,du\end{aligned}$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&= \dfrac13 \int \dfrac{1}{\ u}\,du\\\\\\ &=\dfrac{1}{3}\ln|u|+C\\\\\\\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = 1 3 ln | u | + C = 1 3 ln | x 3 + 4 | + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=\dfrac13\ln|u|+C\\\\\\\ &=\dfrac13\ln|x^3+4|+C\end{aligned} The answer: $\int \dfrac{x^2}{x^3+4}\,dx\,= \dfrac13\ln|x^3+4|+C$